Question

If a 60 ml of water contains 12% of chlorine, how much water must be added in order to create a 8% chlorine solution?​

Answer 1

ɢɪᴠᴇɴ :

• total quantity of water is 60 ml
• contains 12% of chlorine

ᴛᴏ ғɪɴᴅ:

• the quantity of water must be added in order to create a 8% chlorine

sᴏʟᴜᴛɪᴏɴ :

Let x ml of chlorine be present in water.

Then, $\sf{\frac{12}{100} = \frac{x}{60}}$

→ x = $\sf{\frac{12}{100} × 60}$

→ x = 7.2 ml

Therefore, 7.2 ml is present in 60 ml of water.

In order for this 7.2 ml to constitute 8% of the solution, we need to add extra water. Let this be y ml.

Then, $\sf{\frac{8}{100} = \frac{7.2}{y}}$

→ y = 90 ml.

So in order to get a 8% chlorine solution, we need to add 90 - 60 = 30 ml of water.

Answer 2

Answer:

ɢɪᴠᴇɴ :

total quantity of water is 60 ml

contains 12% of chlorine

ᴛᴏ ғɪɴᴅ:

the quantity of water must be added in order to create a 8% chlorine

sᴏʟᴜᴛɪᴏɴ :

Let x ml of chlorine be present in water.

Then, \sf{\frac{12}{100} = \frac{x}{60}}10012=60x

→ x = \sf{\frac{12}{100} × 60}10012×60

→ x = 7.2 ml

Therefore, 7.2 ml is present in 60 ml of water.

In order for this 7.2 ml to constitute 8% of the solution, we need to add extra water. Let this be y ml.

Then, \sf{\frac{8}{100} = \frac{7.2}{y}}1008=y7.2

→ y = 90 ml.

So in order to get a 8% chlorine solution, we need to add 90 - 60 = 30 ml of water.